Remainder Estimate for Integral test

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I have the following question, it is a fill in the blank type question, however when I submit my answer, the system which verifies it say it is incorrect. I believe I am right, so I was hoping for someone to show me where I am going wrong.

Here is the question.

Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n \le$ ____.

Since $R_n \le \int_{n}^\infty \frac{4dx}{x^3} = \lim_{t\to\infty} \int_{n}^t \frac{4dx}{x^3} = $_____.

We want ____ $\le $ _____.

Solving this inequality we get $ n \ge$ ____ $\approx$ ______.

Here is my solution

Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n \le$ 0.0005.

Since $R_n \le \int_{n}^\infty \frac{4dx}{x^3} = \lim_{t\to\infty} \int_{n}^t \frac{4dx}{x^3} = \frac{4}{2n^2}$.

We want $\frac{4}{2n^2}\le 0.0005$

Solving this inequality we get $ n \ge \sqrt{\frac{4}{2\cdot0.0005}} \approx$ 63.24555.

Some other notes... I am pretty convinced that all the blanks are correct except the last blank, thus I have tried 64 as another answer (however that was also flagged as incorrect).